When do two positive braids represent the same link?

Let $B_n$ be the braid group on $n$ strands, with the usual generators: $s_1, \ldots, s_{n-1}$ and their inverses, where $s_i$ is a positive half-twist interchanging the strands labelled $i$ and $i+1$. Markov's theorem says that two elements of the braid group have closures giving the same knot in $S^3$ when they are related by a sequence of moves, each of which is either a conjugation $\beta \to s_i^{\pm} \beta s_i^{\mp}$ or the inclusion $B_n \to B_{n+1}$ taking $\beta \to \beta s_n^{\pm}$.

Recall a braid is positive if it can be written as $s_{i_1} \cdots s_{i_k}$. Conjugation does not generally preserve positive braids, but it can: $s_1^{-1} (s_1 \beta) s_1$ is of course positive when $\beta$ is.

If two positive braids represent the same knot, can they be related by a sequence of Markov moves in which only positive braids appear?

• The paper arxiv.org/pdf/math/0312209.pdf seems relevant, but they only deal with positive permutation braids which close to knots. Commented May 14, 2013 at 6:04
• What you are seeking could be regarded as an analogue of the Tait flyping conjecture for positive braids. Commented May 14, 2013 at 15:55

I'll make some comments on this question.

First of all, two positive braids which are conjugate (equivalently represented by the same labelled oriented 2-component link when taken with the braid axis) will be equivalent by a sequence of positive braid type III Reidemeister moves. This follows from Garside's solution to the conjugacy problem in the braid group (see also Theorem 9.4.2 by Thurston).

For closures of 3-strand braids, the complete classification was given by Birman-Menasco. They show that most 3-strand braids which represent the same link are conjugate, except for braids which are braid index 1 or 2, and a special class of braids $s_1^ps_2^qs_1^rs_2$, where $p,q,r \geq 2$ (restricting to the positive case) which are equivalent to $s_1^ps_2s_1^rs_2^q$. One sees directly from their classification that the braid index 1 and 2 cases have a unique positive representative up to conjugacy. For the exceptional case, they show (or refer to a result of Fein which is given on p. 100 of Birman's book) that these exceptional cases are conjugate after a single stabilization. In fact, one may check directly that the stabilization may be taken to be positive, so this answers your question in the case of pairs of positive 3-strand braids. Also, note that these two examples are related by flypes, and the equivalence by one positive stabilization holds for any pair of braids which are locally related by the same type of flype given in Figure 1.2 (but which are not necessarily conjugate).

If the answer to your question is true, then I think it might give an efficient algorithm to test if two positive braids are equivalent. For positive braids, the Seifert genus is determined by the braid index and number of crossings. When one does several positive Markov stabilizations, one sees that in the positive conjugacy class of the stabilized braid, most braid generators will occur only once. So I think one ought to be able to get a bound on the number of stabilizations needed, which would then lead to an algorithm to tell them apart. One could try to apply the techniques of Birman-Menasco (after Bennequin) to attempt to understand your conjecture.

• Ian, your first paragraph is a bit ambiguous, and the interpretation that was most obvious to me is wrong. It is not true that two conjugate, positive braids are related by cyclic shifts and type III Reidemeister moves. That Theorem you refer to is about conjugation by certain special positive braids, but it's not obvious when you can do that and still have a positive braid, as indeed the "Word Processing" book mentions on the next page (p. 204): $s_2^2 s_1^3$ is conjugate to $s_2^3 s_1^2$, but not by a cyclic shift. Commented Apr 17, 2018 at 3:27
• @DylanThurston do you think your example serves as a counterexample to Vivek's question as stated? As stated, the question allows positive stabilizations and conjugation by a single generator. Conjugation by a single generator is slightly more general than cyclic shifts, but it is still impossible to relate $s_2^2 s_1^3$ to $s_2^3 s_1^2$. But maybe one can still relate them after some positive stabilizations? Commented Aug 3, 2020 at 14:16
• @DylanThurston oops, in fact they can be related by conjugations by single generators: $s_2^2 s_1^3$ -> $s_2 s_1^3 s_2$ -> $s_1 s_2 s_1^3 s_2 s_1^{-1}=s_2^2 s_1^2 s_2$ -> $s_2^3 s_1^2$. Still maybe there is another counterexample. Commented Aug 3, 2020 at 14:23

This was proven in the Paper "Fibered Transverse Knots and the Bennequin Bound" by John B. Etnyre, Jeremy Van Horn-Morris (https://arxiv.org/abs/0803.0758):

Corollary 1.8. Any two positive braids representing a link L are related by positive Markov moves and braid isotopy.

• Why is it "braid isotopy" instead of "braid conjugation"? Commented Aug 3, 2020 at 13:55

The answer is "No". Consider two braids

$$\beta_1 = s_1^3 s_2^3 s_3,\qquad \beta_2 = s_1 s_2^3 s_3^3$$

They obviously represent the same knot. Moreover, they are conjugate. But

Claim 1: $$\beta_2$$ cannot be transformed to $$\beta_1$$ by successive conjugations by $$s_i^{\pm 1}$$ preserving positivity. I.e. you need to conjugate by longer braids. This I checked on a computer. Since there are only finitely many positive braids of given length this is a finite calculation.

Claim 2: $$\beta_2$$ cannot be transformed to $$\beta_1$$ by a sequence of Markov moves preserving positivity. This is conjectural. I checked it on a computer by going through all possible stabilizations with up to 10 strands.

On the other hand, if you allow conjugations by arbitrary short braids (=lifts of permutations) instead of only conjugations by Artin generators, the statement is true by the above cited paper "Fibered Transverse Knots and the Bennequin Bound" by John B. Etnyre, Jeremy Van Horn-Morris (https://arxiv.org/abs/0803.0758) and Lemma 6.7 from Kassel, Turaev - "Braid Groups" (thanks to Vera Vertesi for the latter reference).